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Find x such that the point (x,10) is 5 units from (-7,7)?
high school algebra, do not understand this problem.
2 Answers
- Anonymous1 decade agoFavorite Answer
The distance formula is d = √[(x - x1)² + (y - y1)²]
Use the distance formula to determine the value of x.
5 = √[(x + 7)² + (10 - 7)²] . . . . . .Substitute and evaluate
5 = √(x² + 14x + 49 + (3)²)
5 = √(x² + 14x + 49 + 9)
5 = √(x² + 14x + 58) . . . . . . . . .Square both sides.
5² = √(x² + 14x + 58)²
25 = x² + 14x + 58 . . . . . . . . . .Subtract both sides by 25.
0 = x² + 14x - 25 + 58
0 = x² + 14x + 33v . . . . . . . . . .Factor.
0 = (x + 11)(x + 3) . . . . . . . . . .Set each factor to zero and solve for x.
x + 11 = 0 and x + 3 = 0
x = {-11,-3}
Hence, x = -11 and -3.
I hope this helps!
Source(s): Knowledge - ?Lv 61 decade ago
(x-(-7))^2 + (10-7)^2 = 25
x^2 + 14x + 49 + 9 = 25
x^2 + 14x + 33 = 0
From quadratic equation, x = -3 and x = -11